Question

Solve the following systems of inequations graphically: 

x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x ≥ 0, y ≥ 0 

Answer 1

Converting the inequations to equations, we obtain:
x + 2y = 40, 3x + y = 30, 4x + 3y = 60

x + 2y = 40:  This line meets the x-axis at (40, 0) and y-axis at (0, 20). Draw a thick line through these points.
We see that the origin (0,0) satisfies the inequation x + 2y\[\leq\]40 

Therefore, the region containing the origin is the solution of the inequality x + 2y \[\leq\] 40

3x + y = 30:  This line meets the x-axis at (10, 0) and y-axis at (0, 30). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation 3x + y\[\geq\] 30 

Therefore, the region that does not contain the origin is the solution of the inequality 3x+ y\[\geq\] 30 

4x + 3y = 60:  This line meets the x-axis at (15, 0) and y-axis at (0, 20). Draw a thick line through these points.
We see that the origin (0, 0) does not satisfy the inequation x + 3y \[\geq\] 60 Therefore, the region that does not contain the origin is the solution of the inequality 

4x + 3y\[\geq\]60

Also, x ≥ 0, y ≥ 0 represents the first quadrant. So, the solution set must be in the first quadrant.

Hence, the solution to the inequalities is the intersection of the above three solutions. Thus, the shaded region represents the solution set of the given set of inequalities.